How do you evaluate $sin ({tan}^{- 1} (\frac{7}{9}))$ $sin(tan^-1(7/9))$ ?

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Answer 1 · 2017-01-04T06:31:11

$sin ({tan}^{- 1} (\frac{7}{9})) = \frac{7}{\sqrt{130}}$ $sin(tan^(-1)(7/9))=7/sqrt130$

Let $θ = {tan}^{- 1} (\frac{7}{9})$ $theta=tan^(-1)(7/9)$

then $tan θ = \frac{7}{9}$ $tantheta=7/9$ or $cot θ = \frac{9}{7}$ $cottheta=9/7$

and hence $sin ({tan}^{- 1} (\frac{7}{9}))$ $sin(tan^(-1)(7/9))$

= $sin θ = \frac{1}{csc θ}$ $sintheta=1/csctheta$

= $\frac{1}{\sqrt{{csc}^{2} θ}}$ $1/sqrt(csc^2theta)$

= $\frac{1}{\sqrt{1 + {cot}^{2} θ}}$ $1/sqrt(1+cot^2theta)$

= $\frac{1}{\sqrt{1 + {(\frac{9}{7})}^{2}}}$ $1/sqrt(1+(9/7)^2)$

= $\frac{1}{\sqrt{1 + \frac{81}{49}}} = \frac{1}{\sqrt{\frac{130}{49}}}$ $1/sqrt(1+81/49)=1/sqrt(130/49$

= $\frac{7}{\sqrt{130}}$ $7/sqrt130$

How do you evaluate sin(tan−1(79)) sin(tan^-1(7/9))?